# RMS & Avg. of waves

1. May 25, 2009

### pjrobertson

1. The problem statement, all variables and given/known data

I need to find the average and RMS voltages for the waves (See attached pictures) but I'm not sure exactly how to do it as I don't know how to get an equation for V(t)

2. Relevant equations
I Know:

VRMS = $$\sqrt{\frac{1}{T}\int V(t)^2 dt}$$ between 0 and T

And for Avg it's just VAVG = $$\frac{1}{T}\int V(t) dt$$ between 0 and T

3. The attempt at a solution

I thought I could find the averages by finding the total area for 1 period, then dividing by 1 period. This works (I think, as I don't have the answers)

Giving for the Triangular wave: $$\frac{T}{3} \times \frac{10}{T} = 3.3V$$

and for the second sawtooth type wave: $$\frac{\frac{1}{2}\frac{T}{3} \times 1}{T} = \frac{1}{6}V$$

Thanks

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2. May 25, 2009

### Redbelly98

Staff Emeritus
Looks good for VAVG.

For VRMS, you'll need to write equations for the waveforms. To do that, you may choose any point to represent t=0.

3. May 25, 2009

### pjrobertson

That's the problem, I don't know what to write the equations as. The triangular wave has 3 functions. One for 0 - $$\frac{T}{3}$$ (upwards slanting triangle) one for 0 - $$\frac{2T}{3}$$ (downwards slanting triangle) and one for the last part of the waveform, which is just V(t) = 0.

What are the waveforms for the first 2 sections? And can they be summed like so?:

VRMS = $$\sqrt{\frac{1}{T}\int V_A(t)^2 dt} + \sqrt{\frac{1}{T}\int V_B(t)^2 dt}$$ the first integral being from 0 to T/3 and the second from T/3 to 2T/3

4. May 25, 2009

### Redbelly98

Staff Emeritus
That's pretty much the idea, except that it would be

$$\sqrt{\frac{1}{T}\int V_A(t)^2 dt \ + \ \frac{1}{T}\int V_B(t)^2 dt}$$

with, as you said, "the first integral being from 0 to T/3 and the second from T/3 to 2T/3"

5. May 25, 2009

### pjrobertson

I still can't seem to figure out the RMS values. The problem I have is determining the functions V(t) for each section of the wave.

I have an exam on this tomorrow so any help would be appreciated!

6. May 25, 2009

### Redbelly98

Staff Emeritus
Note that the function is composed of straight line segments. Do you remember, from high school algebra,

y = mx + b

where m is the slope and b is the y-intercept?

For example . . .

. . . for the triangular wave, and 0 ≤ t ≤ T/3:

v(t) = m t + b

What is the slope m in this interval (0 ≤ t ≤ T/3)?
What is the intercept b in this interval?

Find m and b in this interval, and you'll have the function.

Do the same for the interval T/3 ≤ t ≤ 2T/3, and you'll have v(t) in that interval as well.

7. May 25, 2009

### pjrobertson

Hmm...
using v(t) = m t + b
I get:
$$V_A(t) = \frac{30t}{T}$$
$$V_B(t) = 20 - \frac{30t}{T}$$

So I tried integrating all of this, then taking the root (long and tiring process!) and I got V_RMS = 4.71V

Here's the integration I did (without the square root)

$$\sqrt{\frac{30^2}{T^3}\int t^2 dt + \frac{1}{T}\int 400t dt - \frac{1}{T^2}\int 600t^2 dt + \frac{1}{T^3}\int 300t^3 dt}$$

With the first being between 0 and T/3 and the last 3 parts being between T/3 and 2T/3
(there's 3 parts because of squaring $$20 - \frac{30t}{T}$$)

I hope this is right!

8. May 25, 2009

### Redbelly98

Staff Emeritus
You're on the right track, but I do see a couple of errors.

You seem to have expanded (20 - 30t/T)2 incorrectly, in particular the "middle" term's coefficient of -600 is wrong.

Also, there is an extra factor of t in 3 of your integrals.